Question

Usually I don't ask serious questions cause I consider myself smart but I'm stuck-
(Algebra 2 easy)
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A certain bacteria colony grows according to: N = $N_{0} e^{0.45t}$, where t is time in hours. At 7 am, there are 400 bacteria. What is the equation for this? How large is the colony at 2 pm the following day?
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Bonus question: What was the colony population 3 hours before the experiment began?

Answer 1

Answer:

N=400×e^(0.45t)

4.58×10^8 at 2pm next day

104 for 3 hours before

(Such a fast rate...wondering if there were more 0's behind that decimal)

Step-by-step explanation:

Let's assume 7am is the intial. So at t=0, we have N_0=400.

This means the equation is N=400×e^(0.45t).

2pm next day is 7+24 hours later. So replace t with 31 giving us N=400×e^(0.45×31)=4.58 ×10^8 approximately.

3 hours before the expression began at at t=-3:

N=400×e^(0.45×-3)=104 approximately.